CisternPipes

Question

A tank is filled by three pipes with uniform flow. The first two pipes operating simultaneously fill the tank in the same time during which the tank is filled by the third pipe alone. The second pipe fills the tank 5 hours faster than the first pipe and 4 hours slower than the third pipe. The time required by the first pipe is:

A.    6 hours    B.    10 hours
C.    15 hours    D.    30 hours

Comments

15

Answer 1

Option c

Let A is 1st pipe B 2nd and C 3rd pipe

Let B can do work in x hr

A  can do work in x+5 hr

C  can do work in x-4 hr

A can fill tank in 1 hr is 1/x+5 work

 B can fill tank in 1 hr is 1/x work

A and B together can fill tank in 1 hr is (1/x) + 1/(x+5)

=(2x+5)/(x²+5x) work

A and B can together fill the tank in (x²+5x)/(2x+5) hr 

(which is equal to the time C cn fill the tank alone)

X-4= (x²+5x)/(2x+5)

By solving it x=10 hr

So A cn fill the tank in 10+5= 15 hr 

Comments

How to get c=x-4?

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Because  it is given in the ques that second pipe(pipe b) is 4 hr slower than 3rd pipe(c pipe).

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Yes got it...thanks

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We can do like this also ⬇

Suppose, first pipe alone takes x hours to fill the tank .

Then, second and third pipes will take (x -5) and (x - 9) hours respectively to fill the tank.

so, (1/x) + (1/x-5) = (1/x-9)